Solve

Asked by ganesh_94 | 22nd Aug, 2009, 08:24: PM

Expert Answer:

Let P(x1,y1) and Q(x2,y2) be the points,

then using distance formula,

AP = PQ = QB

AP2 = PQ2 = QB2

(y1-1)2+(x1-2)2 = (y2-y1)2+(x2-x1)2 = (-8-y2)2+(5-x2)2  

PQ2 = QB2

(y2-y1)2+(x2-x1)2 = (-8-y2)2+(5-x2)2 

y22 - 2y1y2 + y12 + x22 - 2x1x2 + x12 =  y2+ 16y2 + 64 + x22 - 10x2 + 25

y22 - 2y1y2 + y12 + x22 - 2x1x2 + x12 =  y2+ 16y2 + 89 + x22 - 10x2

Comparing the coefficients of y2 and x2, on both sides,

-2y1 = 16 and -2x1 = -10

Hence, y1 = -8, x1 = 5,

P(5,-8), and it lies on the line 2x-y-k = 0,

2(5)-(-8) = k

k = 18

Regards,

Team,

TopperLearning.

 

Answered by  | 24th Aug, 2009, 09:36: AM

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